Past exam of the soft condensed matter course of the University of Cambridge / 2021 / Section B / 6 / c / Solution

At the thermodynamic critical point, the two spinodal branches merge: $f^{\prime \prime }=0$ and $f^{(3)}=0$. The third derivative isFor $0<\varphi <1$, its zero gives $1-{\varphi }_c=\sqrt{N}{\varphi }_c$. Substitution in the spinodal relation yieldsAn absolute critical temperature requires an interaction–temperature relation. Under the usual energetic approximation $\chi =C/T$, define the theta temperature $\Theta =2C$, at which an infinite chain has $\chi =1/2$. ThenThus $T_c$ tends to a nonzero constant, $T_c\sim N^0$, while $\Theta -T_c\sim N^{-1/2}$. It is the temperature deficit, not $T_c$ itself, that scales as $N^{-1/2}$.

More generally, if a smooth $\chi (T)$ obeys $\chi (\Theta )=1/2$ and ${\chi }^{\prime }(\Theta )\ne 0$, thenThis is the critical point of a Flory–Huggins polymer solution. The same leading chain-length exponent follows for a smooth temperature law, while its coefficient depends on the omitted solvent thermodynamics. The lattice interpretation connects the chain-length dependence of mixing entropy to the theta condition $\chi =1/2$.^[ref].